# Ratio test for series

woopydalan
Homework Statement:
Use the ratio test for the series ##\sum_{n=1}^{\infty} b_{n}## where ##b_{1} = 5## and ##b_{n}= \frac {(-1)^{n}nb_{n-1}}{3n+1}## for ##n \geq 2##
Relevant Equations:
ratio test: ##L = \lim_{n \to \infty} \lvert \frac {a_{n+1}}{a_{n}} \rvert## will converge if ##L < 1## and will diverge if ##L > 1##
So I am having some difficulty expressing this series explicitly. I just tried finding some terms

##b_{0} = 5##

I am assuming I am allowed to use that for ##b_{1}## for the series, even if the series begins at ##n=1##? With that assumption, I have

##b_{1} = -\frac {5}{4}##
##b_{2} = - \frac{5}{14}##
##b_{3} = \frac {3}{28}##
## b_{4} = \frac {3}{91}##

I am not seeing a pattern. I also alternatively tried using the ratio test not in explicit terms

##\lim_{n \to \infty} \lvert \frac {(-1)^{n+1}(n+1)b_{n}}{3(n+1)+1} \cdot \frac {3n+1}{(-1)^{n}nb_{n-1}} \rvert## and after substituting I ended up with ##\lvert \frac{b_{n}}{b_{n-1}} \rvert## which to me is inconclusive.

Last edited:

Staff Emeritus
Homework Helper
Homework Statement:: Use the ratio test for the series ##\sum_{n=1}^{\infty} b_{n}## where ##b_{1} = 5## and ##b_{n}= \frac {(-1)^{n}nb_{n-1}}{3n+1}## for ##n \geq 2##
Relevant Equations:: ratio test: ##L = \lim_{n \to \infty} \lvert \frac {a_{n+1}}{a_{n}} \rvert## will converge if ##L < 1## and will diverge if ##L > 1##

So I am having some difficulty expressing this series explicitly. I just tried finding some terms
You're making this way too complicated.
$$b_{n}= \frac {(-1)^{n}nb_{n-1}}{3n+1} \quad \Rightarrow \quad \frac{b_{n}}{b_{n-1}}= \frac {(-1)^{n}n}{3n+1}$$

woopydalan
woopydalan
Oh wow...let me try that

woopydalan
Ok, so now I got ##\lim_{n \to \infty} \frac {n}{3n+1}## which would be indeterminate??

Staff Emeritus
Homework Helper
$$\lim_{n \to \infty} \frac{an+b}{cn+d} = \lim_{n \to \infty} \frac{a + \frac{b}{n}}{c + \frac{d}{n}} = \frac{a}{c}$$ for $c \neq 0$.